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ANANDU

· started a discussion

· 1 Months ago

ABCD is square right? diagonal of ABCD= a/root2
therefore radius of circle= a/2root2
comparing circumradius of equilateral triangle, we get side of eq triangle= root3*a/2root2.
so ANS should have been 3root3a^2/32

Question:
What is the area of the inner equilateral triangle if the side of the outermost square is ‘a’? (ABCD is a square).

Options:
A) \(\cfrac{3\sqrt[]{3}a^2}{32}\)
B) \(\cfrac{\sqrt[]{3}a^2}{16}\)
C) \(\cfrac{5\sqrt[]{3}a^2}{32}\)
D) \(\cfrac{5\sqrt[]{3}a^2}{64}\)
Solution:
Ans: (b) \(BD=a,EF=\cfrac{a}{2}\)

\(\therefore\) area of equilateral triangle EFG \(=\cfrac{\sqrt[]{3}}{4}\left ( \cfrac {a}{2} \right )^2=\cfrac{\sqrt[]{3}a^2}{16}\)

Knowledge Expert

· commented

· 1 Months ago

@ANANDU Remember, you have to calculate area of equilateral triangle.

ANANDU

· commented

· 1 Months ago

side of ABCD=a/root2

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